Mathematics 2865 Assignment #2
Solutions
1) Find the number of permutations that contain either bge, eaf or both and subtract
from the total number of permutations (7!). Let A be the set of permutations
which contain the pattern bge: |A| = 5! (thinking of bge as a single symbol).
Similarly, if B is the set of permutations which contain eaf, |B| = 5!. For
permutations to contain both bge and eaf it must contain bgeaf, so |A∩B| = 3!
The number containing either bge or eaf is |A∪B| = |A| + |B| - |A∩B| = 5! + 5! –
3! = 234. Hence, the answer is 7! – 234 = 5040 – 234 = 4806
2) Use addition rule. First assume there are two letters between a and b. The
number of permutations here is 2*P(5,2)*4! = 960 (there are P(5,2) permutations
of 2 letters that don’t include a and b, then 2 ways to place a and b on each end
and finally glue those 4 letters together and arrange with the rest of the letters in
4! ways). Next assume there are three letters between a and b. The number of
such permutations is 2*P(5,3)*3! = 720. By the addition rule, the answer is 960 +
720 = 1680
3) Order doesn’t matter here!
"9%
a) $ ' = 84
#6&
" 7%
b) The number of groups that include divorced couple is $ ' (couple is already
# 4&
invited, so choose the other 4 guests out of the remaining 7). Thus, the number in
!
"9% " 7%
which not both of these people are present is $ ' ( $ ' = 84 ( 35 = 49
#6!& # 4 &
c) Our hostess must either invite two married couples and two singles or all three
married couples. If she invites two married couples and two singles the answer is
" 3%" 3%
couples the answer is 1 since there is
!
$ '$ ' and if she invites all three married
#2&# 2&
" 3%" 3%
only one way to choose 3 couples out of 3. Now use addition rule, $ '$ ' + 1 = 10
#2&# 2&
!
!
!
4) Think of the coins as being indistinguishable object. Now place them into
distinguishable bins named pennies, nickels, dimes and quarters. There are
!
#10 + 4 "1&
%
( ways to do that, or 286.
$ 10 '
5) The general term is
"58%" (3 % 58(k
"58%
"58%
k
58(k
(2 58(k k k
58(k k (116+3k
$ '$ 2 ' (2x) = $ '((3) (x ) 2 x = $ '((3) 2 x
#
&
k
k
k
x
# &
# &
# &
Now we want the coefficient of x25, therefore -116+3k must equal 25. It follows
that k = 47. Plug in the value for k in the coefficient above and the answer is
" 58%
11 47
$ '((3) 2
# 47&
6) 0 _____ 1 _____ 0 _____ 1 _____ 0 _____ this is the sequence. Here we
have 5 bins and 20 – 5 objects to distribute. Consider these 15 0s and 1s as
indistinguishable objects being placed in distinguishable bins. Therefore, there
#15 + 5 "1&
are %
( = 3876 such sequences.
$ 15 '
!
!
!
"2n % " n %
7) $ ' = 2$ ' + n 2 Left hand side: Selecting 2 people out of n men and n
# 2 & # 2&
"2n %
women. There is $ ' ways to do that. Right hand side: Same selection in a
#2&
different way. We can choose two men, or two women or one of each. If it’s
" n%
" n%
men, then number of ways is $ ' , if it’s women the number of ways is $ ' and if
# 2&
# 2&
!
it’s one of each then the number of ways is n*n, since there are n choices for men
and n for women. Use addition rule and the result follows.
!
" n % " n% " n %
n!
n!
n!
+6
+6
$ ' + 6$ ' + 6$ ' =
2!(n ( 2)!
3!(n ( 3)!
8) # 1 & # 2 & # 3& 1!(n (1)!
!
= n + 3n(n (1) + n(n (1)(n ( 2) = n + 3n 2 ( 3n + n 3 ( 3n 2 + 2n = n 3
!